If n(A)=43, n(B)=51 and n(A∪ B)=75, then n(A-B)∪(B-A) is equal to

Question

If n(A)=43, n(B)=51 and n(A∪ B)=75, then n(A-B)∪(B-A) is equal to (A) 53 (B) 45 (C) 56 (D) 66

Tardigrade Admin · Accepted Answer

Given, n(A)=43, n(B)=51 and n(A ∪ B)=75 Now, by addition theorem of probability, n(A ∩ B) =n(A)+n(B)-n(A ∪ B) =43+51-75=19 Now, n[(A-B) ∪(B-A)] =n(A ∪ B)-n(A ∩ B) =75-19=56

Q. If $n (A) = 43, n (B) = 51 an d n (A \cup B) = 75, t h e n n (A - B) \cup (B - A)$ is equal to

$53$

$45$

$56$

$66$

$46$

Given, $n (A) = 43, n (B) = 51$ and $n (A \cup B) = 75$
Now, by addition theorem of probability,
$n (A \cap B) = n (A) + n (B) - n (A \cup B)$
$= 43 + 51 - 75 = 19$
Now, $n [(A - B) \cup (B - A)]$
$= n (A \cup B) - n (A \cap B)$
$= 75 - 19 = 56$

Q. If n(A)=43,n(B)=51andn(A∪B)=75,thenn(A−B)∪(B−A) is equal to

53

45

56

66

46

Given, n(A)=43,n(B)=51 and n(A∪B)=75 Now, by addition theorem of probability, n(A∩B)=n(A)+n(B)−n(A∪B) =43+51−75=19 Now, n[(A−B)∪(B−A)] =n(A∪B)−n(A∩B) =75−19=56